Question 883955
{{{graph(300,300,-2,2,-2,2,4x^3+14x-8 )}}}
Looks like the zero lies between 0 and 1.
I used Newton's method with a starting value of 0.5.
{{{x[n+1]=x[n]-f(x[n])/((df/dx)(x[n]))}}}
with
{{{f=4x^3+14x-8}}}
{{{df/dx=12x^2+14}}}
{{{x=0.529107}}}